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Finite Fields and Their Applications 18 2012 791 799 Contents lists available at SciVerse ScienceDirect Finite Fields and Their Applications www.elsevier.

Chandler b, Junhua Wu a,2, Qing Xiang b,,1 a Department of Mathematics, Lane College, Jackson, TN 38301, USA b Department of Mathematical Sciences, University of Delaware, Newark, DE 19716, USA

2 Finite Fields and Their Applications Contents lists available at SciVerse ScienceDirect Finite Fields and Their Applications Power sums over subspaces of finite fields David B. Chandler b, Junhua Wu a,2, Qing Xiang b,,1 a Department of Mathematics, Lane College, Jackson, TN 38301, USA b Department of Mathematical Sciences, University of Delaware, Newark, DE 19716, USA article info abstract Article history: Received 31 August 2011 Revised 16 December 2011 Accepted 11 April 2012 Available online 25 April 2012 Communicated by L. Storme MSC: 05B25 11T24 11T71 Let K be the finite field of order q m+1, which is regarded as an m + 1-dimensional vector space over F q.foreachh-dimensional F q -subspace V of K, α K and 0 t q m+1 1, we define S t V,α = v V α + vt.foreach1 h m, weobtainsufficient conditions on t for the vanishing of S t V,α; when h = m, combining this result with some p-rank results from coding theory, we obtain necessary and sufficient conditions on t for the vanishing of S t V,α Elsevier Inc. All rights reserved. Keywords: Code p-rank Power sum Rank 1. Introduction Let K be the finite field of order q m+1, where q = p e is a prime power and m 0. It is a wellknown and very useful fact that v K { v t 0, if t 0 mod q = m+1 1 or t = 0, 1, if t 0 mod q m+1 1 and t > 0. * Corresponding author. addresses: davidbchandler@gmail.com D.B. Chandler, jwu@lanecollege.edu J. Wu, xiang@math.udel.edu Q. Xiang. 1 Research supported in part by NSF Grant DMS Research supported in part by NSF HBCU-UP Grant Award at Lane College /$ see front matter 2012 Elsevier Inc. All rights reserved.

3 792 D.B. Chandler et al. / Finite Fields and Their Applications Here we have used the convention that 0 0 = 1. In connection to this fact, the following question arises naturally. If we restrict the range of the above summation to some subset of K instead of K itself, is it possible to determine for which t the sum vanishes? Probably the most natural subsets to consider are the F q -subspaces of K, which is now viewed as an m + 1-dimensional vector space over F q. Let V be a nonzero F q -subspace of K.Foreach0 t q m+1 1 and α K,wedefine S t V,α = v V v + α t. 1.1 These sums arose in an investigation of the integral Galois module structure of certain extensions of p-adic fields [4]. They also arose, at least in some special cases, in the study of zeros and nonzeros of certain cyclic codes from finite geometry. In the case where e = 1 i.e., q = p is a prime and V is an m-dimensional subspace of K, Byott and Chapman [5] give necessary and sufficient conditions for the vanishing of S t V, α in terms of the digit sum of t. In the same paper, the authors also collect some conditions on t under which the sum in 1.1 vanishes in the case where e 1 and V is an m-dimensional F q -subspace of K. In this paper we consider the problem of vanishing or nonvanishing of the sum S t V, α in the general case where e 1 and V is an arbitrary nonzero F q -subspace of K.WhenV is an arbitrary nonzero F q -subspace of K, we give sufficient conditions on t for the vanishing of the sum in 1.1. Moreover, when V is an m-dimensional F q -subspace of K, we give necessary and sufficient conditions for the vanishing of the sum in 1.1. It is clear that S q m+1 1 V, α = 1 or 0 according as α V or α / V. So we only need to consider those t such that 0 t q m+1 2. To state our theorem, we need some notation. For 0 t q m+1 2, let t = m t iq i be the base-q expansion of t, where 0 t i. For each 1 l e, wedefine s l t := 1 m σ q p e l ti, 1.2 where σ q x is the base-q digit sum of x. We remark that the e-tuples s 1 t, s 2 t,...,s e t are similar to the twisted degrees which appeared in [6] and [3]. For future use we also denote by σ x the base-p digit sum of x, where x is a nonnegative integer. Now we can state our main result. Theorem 1.1. Let K be the finite field of order q m+1,wherem 0 and q = p e is a prime power with e 1.Let 1 h m, and V be an h-dimensional F q -subspace of K. For 0 t q m+1 2 and α K, let S t V, α be the sum defined in 1.1. i When α V,wehave S t V, α = 0 if tors l t<hforsome1 l e. When α / V,wehave S t V, α = 0 if s l t<hforsome1 l e. ii Assume that h = m. Let α V.Then S t V, α = 0 if and only if either tors l t<mforsome 1 l e. Suppose now α / V.Then S t V, α = 0 if and only if s l t<mforsome1 l e. We remark that our approach to the proof of Theorem 1.1 is completely different from the one used in [5]. The approach in [5] started by considering the generating function of S t V, α t 0. Therefore it can be viewed as external. Our approach is internal in the sense that we try to expand the sum S t V, α in a certain way. When e = 1, i.e., q = p is a prime, if V is an m-dimensional F q -subspace of K, using ii of Theorem 1.1 and noting that S p m+1 1 V, α = 1 or 0 according as α V or α / V, we have the following corollary, which is one of the main results in [5]. Corollary 1.2. See [5, Theorem 2]. Let K = F p m+1, let V be an m-dimensional F p -subspace of K, and let 0 t p m+1 1.

4 D.B. Chandler et al. / Finite Fields and Their Applications i Suppose α K \ V.Then S t V, α 0 if and only if p 1m σ t<p 1m + 1. ii Suppose α V.Then S t V, α 0 if and only if σ t = p 1m ort= p m+1 1. We give a brief overview of the paper. In Section 2, we give the proof of part i of Theorem 1.1. We expand S t V, α and show that if the conditions of the theorem are met, then every term in the expansion vanishes. In Section 3, we give the proof of part ii of Theorem 1.1. The proof is done by combining the result in part i of Theorem 1.1 and some p-rank results from coding theory. 2. Proof of the first part of Theorem 1.1 In this section we give a proof of Theorem 1.1i. The ideas of the proof are similar to those in the proof of Wan s Theorem [7, Theorem 1.3]. Proof of Theorem 1.1i. Let t = t 0 + t 1 q + +t m q m be the base-q expansion of t, and let {x 1, x 2,...,x h } be an F q -basis of V.Wehave S t V,α = = r 1,r 2,...,r h F q r 1 x 1 + r 2 x 2 + +r h x h + α t r 1,r 2,...,r h F q m r1 x qi 1 + r 2x qi 2 + +r hx qi h + ti αqi because elements of F q are fixed when raised to the power q. Then S t V,α = t i =k i,1 +k i,2 + +k i,h+1 0 i m h m r k i, j j r j F q r σ r p 1 m t i k i,1,k i,2,...,k i,h+1 h m x qi k i, j j α m qi k i,h Using Legendre s formula, ν p r! = here ν p is the p-adic valuation function, we find that the p-adic valuation of the product of the multinomial coefficients in 2.1 is 1 p 1 m h+1 t i σ t i ki, j σ k i, j = 1 p 1 m h+1 σ k i, j σ t i. 2.2 We will show that if s l t<h for some l, 1 l e, then the quantity in the right-hand side of 2.2 is positive, implying that the corresponding term in the expansion 2.1 of S t V, α is zero. We start by noting that { 0, if does not divide r, x r = 0, if r = 0, x F q 1, if r and r > Therefore in 2.1 we only need to consider those terms for which m k i, j 0 mod and m k i, j > 0 2.4

5 794 D.B. Chandler et al. / Finite Fields and Their Applications for all j = 1, 2,...,h. Sincer σ q r mod, we see that the congruence in 2.4 implies m σ q k i, j 0 mod 2.5 for all j = 1, 2,...,h. Given nonnegative integers k i, j such that h+1 k i, j = t i for 0 i m and 2.4 is satisfied for each j, 1 j h, we consider the following vector of length h + 1 m m m k i,1, k i,2,..., k i,h The first h entries of the vector in 2.6 are nonzero by assumption. It follows that the corresponding entries of the following vector are at least. Therefore we have m m m σ q k i,1, σ q k i,2,..., σ q k i,h+1 h m t i m h σ q k i, j t i. 2.8 For any nonnegative integer l, we note that 2.5 still holds with σ q k i, j replaced by σ q p l k i, j.also m σ qp l k i, j is still nonzero for the first h subscripts of j. Thuswehave h m m σ q p l ti h+1 σ q p l ki, j σq p l ti. 2.9 Since h+1 k i, j = t i the right-hand side of the above inequality is nonnegative; we sum over l to get { e 1 max 0, h l=0 m σ q p l ti } p 1 m σ k i, j σ t i h+1 Here we have used the fact that e 1 σ l=0 qp l r = q 1 σ p 1 r. Recall that by definition 1.2, s l = 1 m σ q 1 qp e l t i. We see that the left-hand side of 2.10 is e 1 { max 0, h s e l t }. l=0 If s l t<h for some 1 l e, then by 2.10, we have m h+1 σ k i, j σ t i > 0;

6 D.B. Chandler et al. / Finite Fields and Their Applications it follows that the corresponding term in the right-hand side of 2.1 is zero. Therefore we have shown that if s l t<h for some 1 l e then S t V, α = 0. Finally when α V,wehaveS t V, α = S t V, 0. Letθ be a primitive element of F q.then S t V, 0 = S t θ V, 0 = θ t S t V, 0. If t 0 mod then θ t 1, and thus S t V, α = 0. The proof of Theorem 1.1i is now complete. 3. Proof of the second part of Theorem 1.1 In this section, we always assume that V is an m-dimensional F q -subspace of K ; it follows that V =q m. In this case, using some results from coding theory, we prove that the conditions in Theorem 1.1i are indeed necessary for the vanishing of 1.1. We start by counting the sizes of two important sets that will be used later. Let N := { t 0 t q m+1 2, s l t m for each 1 l e } 3.1 and N := { t t N and t }. 3.2 Lemma 3.1. Using the above notation, we have m + e p N = 1 and N m + p e 1 =. m + 1 m Proof. Let t = m t iq i be the base-q expansion of t. It is clear that s l q m+1 1 = m + 1 and s l q m+1 1 t = 1 = 1 m σ q p e l t i m = m + 1 s l t σ q p e l 1 m σ q p e l ti for each 1 l e. This implies that s l t m if and only if s l q m+1 1 t 1. Consequently the sets M := { t 1 t q m+1 1, s l t 1foreach1 l e } and M := { t t M and t } are of the same sizes as the sets N and N, respectively. The rest is to count the sizes of M and M.

7 796 D.B. Chandler et al. / Finite Fields and Their Applications Let t M, t = m t iq i,0 t i, i, and let t i = t i,0 + t i,1 p + +t i,e 1 p e 1 be the base-p expansion of t i for 0 i m, where 0 t i, j p 1. If for some 0 k e 1, then m t i,k p s k+1 t = 1 1 m σ q p e 1 k t i m t i,k p e 1 p pe 1 > 1, which is not the case by the choice of t. Therefore we must have m t i, j p 1, 0 j e On the other hand, if 3.3 is satisfied and t 0, then t M. Thus, M isonelessthanthenumber of ways to distribute p 1 indistinguishable balls into m + 2 labeled boxes, e consecutive times. The first result follows. If 3.3 is satisfied but the inequality is strict for some j, thenσ q t<q 1, which means t / M since σ q t t mod. Hence to get the second result we distribute p 1 indistinguishable balls into only m + 1 labeled boxes, e consecutive times. The proof of the lemma is now complete. Recall that K is the finite field of order q m+1.intherestofthearticle,wewillalwaysuseg to denote the multiplicative group K of K ; that is, G = ξ is a cyclic group of order q m+1 1, where ξ is a primitive element of K. It is well known that all the irreducible K -characters of G are linear and form the following set Ĝ = { χ i 0 i q m+1 2 }, where χ i ξ j = ξ ji ;inparticular,χ 0 is the trivial character of G. Let B = g G b g g be an element in the group algebra K [G]. We associate with B a square matrix b h 1 g whose rows and columns are indexed by h G and g G, respectively. The rank of b h 1 g over K is defined to be the rank of B. The following result is well known. Lemma 3.2. See [2, Theorem 3.3]. Let B = g G b g g K [G]. Then the rank of B is equal to the number of K -characters χ ĜsuchthatχB = g G b gχg 0. Now we remind the reader that V is an m-dimensional F q -subspace of K ; that is, V is a hyperplane through the origin of K regarded as an F q -vector space. For convenience, we define the following two sets:

8 D.B. Chandler et al. / Finite Fields and Their Applications H := {U U is a codimension-1 subspace of K} 3.4 and H := {r + U U H, r / U}, 3.5 where r + U ={r + u u U}. Letg G. WedefinegU ={gu u U} and gr + U ={gr + gu u U}. Then G acts on each of H and H. Lemma 3.3. Using the above notation, G acts transitively on each of H and H. Proof. Let Tr q m+1 /q : F q m+1 F q be the trace map. Every hyperplane of K through the origin has the form {β: Tr q m+1 /qgβ = 0} for some g G. Every hyperplane of K not through the origin has the form {β: Tr q m+1 /q gβ = 1} for some g G. ThusG acts transitively on each of H and H. Let 1 K denote the all-1 row vector indexed by K. We define the following matrices: A is the incidence matrix between the elements of H and G; that is, A is a 0, 1-matrix whose rows and columns are labeled by the elements of H and G, respectively; the U, g-entry of A is 1 if and only if g U. B is the incidence matrix between the elements of H H and K. C is the incidence matrix between the elements of H and K. C is the matrix obtained from C by deleting the column of C indexed by 0. H is obtained by adjoining to C the row 1 K. As defined, B is the incidence matrix between the hyperplanes and the points of AGm + 1, q the affine space of dimension m + 1 whose hyperplanes are the cosets of subspaces of dimension m and whose points are the element of K. Then C is the incidence matrix between the hyperplanes not through the origin and the points of AGm + 1, q. Thep-rank of a matrix is its rank over any field of characteristic p which contains all the entries of the matrix. For a matrix E with entries in K,welet rk p E denote the p-rank of the matrix E. Lemma 3.4. Using the above notation, we have i rk p A = m+p 1 e m + 1; ii rk p B = m+p e; m+1 iii rk p C = rk p C = m+p e m+1 1. Proof. Let D be the incidence matrix of the hyperplanes and points of PGm, q, thefiniteprojective space of dimension m. By [1, Theorem 5.7.1], we have that rk p D = m+p 1 e m + 1. Deleting the duplicated columns of A, weobtaind. Thusrk p A = rk p D = m+p 1 e m + 1. By [1, Corollary 5.7.1], it follows that rk p B = m+p e. m+1 It remains to establish iii. First we note that all the entries in the column of C indexed by 0 are zero since no element in H contains 0. Therefore, rk p C = rk p C. LetB W denote the row of B indexed by W H H.Since B U + V CU B V = 1 K for each U H, where CU ={r + U r K, r U}, it is clear that the row span of H is the same as that of B. Since the entries of the column of C indexed by 0 are all zero, 1 K is not contained in the

9 798 D.B. Chandler et al. / Finite Fields and Their Applications span of the rows of C. We have proved that rk p C m + e p = rk p C = rk p B 1 = rk p H 1 = 1. m + 1 Proof of Theorem 1.1ii. Let V K be an F q -subspace of dimension m. We first consider the sums S t V, α with α V. In this case, we clearly have S t V, α = S t V, 0. Define and Z := { t 0 t q m+1 2, s l t<m for some 1 l e or t } Z 1 := { t 0 t q m+1 2, S t V, 0 = 0 }. By part i of Theorem 1.1, we have Z Z 1. From Lemma 3.1, we have Z = q m+1 1 N = q m+1 1 m + p e 1, 3.6 where N is the set defined in 3.2. Consider B = v V v K [G], where V = V \{0}. SinceG is transitive on H by Lemma 3.3, it is clear that, up to permutations of rows and columns and repetition of rows, the matrix associated with B is equal to A. By Lemma 3.2, we know that the number of characters in Ĝ ={χ t 0 t q m+1 2} satisfying χ t B = v V χ t v = v V v t 0 m+p 1 e is equal to rk p A, which in turn is equal to m + 1 by Lemma 3.4. Furthermore, since when t = 0, χ 0 B = q m 1 0but v V v0 = q m = 0, we conclude that the number of t with 0 t q m+1 2 satisfying m S t V, 0 = v V v t 0 m+p 1 e, is m and Z1 =q m+1 1 m+p 1 e. m With 3.6 and Z Z 1, we obtain that Z = Z 1.The proof of the first part of Theorem 1.1ii is complete. Next let α / V.Wedefine and Z := { t 0 t q m+1 2, s l t<m for some 1 l e } Z 2 := { t 0 t q m+1 2, S t V,α = 0 }. By part i of Theorem 1.1, we have Z Z 2. From Lemma 3.1, it follows that m + e p Z =q m+1 1 N =q m+1, 3.7 m + 1

10 D.B. Chandler et al. / Finite Fields and Their Applications where N is the set defined in 3.1. Let B K [G] be the element corresponding to the subset {α + v v V } of G. That is, B = g G b g g, where b g = 1ifg {α + v v V } and b g = 0otherwise.Since G is transitive on H by Lemma 3.3, it is clear that, up to permutations of rows and columns, the matrix associated with B is equal to C. By Lemma 3.2, we know that the number of characters in Ĝ ={χ t 0 t q m+1 2} satisfying χ t B = v V χ t α + v = v V α + v t 0 is equal to rk p C, which by Lemma 3.4iii in turn is equal to m+p m+1 e 1. Consequently, the number of t with 0 t q m+1 2 satisfying S t V,α = v V α + v t = 0 equals q m+1 m+p e,orequivalently, Z2 =q m+1 m+p e. m+1 m+1 With 3.7 and Z Z2, we obtain that Z = Z 2. The proof of the second part of Theorem 1.1ii is now complete. Remark 3.5. It is natural to ask whether one can obtain necessary and sufficient conditions on t such as those in part ii of Theorem 1.1 for the vanishing of S t V, α when the dimension of V is less than m. We believe that the answer to this question is negative. The reason is that while G acts transitively on H, it does not act transitively on the set of h-dimensional F q -subspaces of K, where 1 < h < m. References [1] E.F. Assmus Jr., J.D. Key, Designs and Their Codes, Cambridge University Press, New York, [2] J. MacWilliams, H.B. Mann, On the p-rank of the design matrix of a difference set, Inform. Control [3] M. Bardoe, P. Sin, The permutation modules for GLn + 1, F q acting on P n F q and F n+1 q, J. Lond. Math. Soc [4] N.P. Byott, Integral Galois module structure of some Lubin Tate extensions, J. Number Theory [5] N.P. Byott, R.J. Chapman, Power sums over finite subspaces of a field, Finite Fields Appl [6] N. Hamada, The rank of the incidence matrix of points and d-flats in finite geometries, J. Sci. Hiroshima Univ. Ser. A-I [7] D. Wan, A Chevalley Warning approach to p-adic estimates of character sums, Proc. Amer. Math. Soc

This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and

This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and education use, including for instruction at the authors institution